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Done two-pointers-2 #1759

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28 changes: 28 additions & 0 deletions problem1.java
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// Time Complexity : O(n)
// Space Complexity : O(1)
// Did this code successfully run on Leetcode : Yes
// Any problem you faced while coding this : No, just needed to allow only 2 duplicates and shift in-place.


// Your code here along with comments explaining your approach in three sentences only
// Since array is sorted, same numbers come together, so we can control how many times we keep them.
// I use a write pointer and allow a number only if it is different from nums[write-2] (means max 2 allowed).
// This updates the array in-place and returns the new length.

class Solution {
public int removeDuplicates(int[] nums) {

if (nums.length <= 2) return nums.length;

int write = 2;

for (int i = 2; i < nums.length; i++) {
if (nums[i] != nums[write - 2]) {
nums[write] = nums[i];
write++;
}
}

return write;
}
}
30 changes: 30 additions & 0 deletions problem2.java
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// Time Complexity : O(m + n)
// Space Complexity : O(1)
// Did this code successfully run on Leetcode : Yes
// Any problem you faced while coding this : No, just needed to merge from end to avoid overwriting nums1.


// Your code here along with comments explaining your approach in three sentences only
// Since nums1 has extra space at the end, I fill it from the back so existing values don't get overwritten.
// I compare nums1[m-1] and nums2[n-1] and place the bigger one at the end pointer.
// Continue until nums2 is fully placed, because remaining nums1 part is already in correct position.

class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {

int i = m - 1;
int j = n - 1;
int k = m + n - 1;

while (j >= 0) {
if (i >= 0 && nums1[i] > nums2[j]) {
nums1[k] = nums1[i];
i--;
} else {
nums1[k] = nums2[j];
j--;
}
k--;
}
}
}
30 changes: 30 additions & 0 deletions problem3.java
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// Time Complexity : O(m + n)
// Space Complexity : O(1)
// Did this code successfully run on Leetcode : Yes
// Any problem you faced while coding this : No, just needed to start from correct corner.


// Your code here along with comments explaining your approach in three sentences only
// I start from the top-right corner because from there I can eliminate one full row or column each step.
// If the current value is bigger than target, I move left, and if smaller, I move down.
// This way I never revisit cells and finish in O(m+n).

class Solution {
public boolean searchMatrix(int[][] matrix, int target) {

int m = matrix.length;
int n = matrix[0].length;

int r = 0;
int c = n - 1;

while (r < m && c >= 0) {

if (matrix[r][c] == target) return true;
else if (matrix[r][c] > target) c--;
else r++;
}

return false;
}
}